0623. Add One Row to Tree (M)

Add One Row to Tree (M)

题目

Given the root of a binary tree, then value v and depth d, you need to add a row of nodes with value v at the given depth d. The root node is at depth 1.

The adding rule is: given a positive integer depth d, for each NOT null tree nodes N in depth d-1, create two tree nodes with value v as N's left subtree root and right subtree root. And N's original left subtree should be the left subtree of the new left subtree root, its original right subtree should be the right subtree of the new right subtree root. If depth d is 1 that means there is no depth d-1 at all, then create a tree node with value v as the new root of the whole original tree, and the original tree is the new root's left subtree.

Example 1:

Input: 
A binary tree as following:
       4
     /   
    2     6
   /    / 
  3   1 5   

v = 1

d = 2

Output: 
       4
      / 
     1   1
    /     
   2       6
  /      / 
 3   1   5   

Example 2:

Input: 
A binary tree as following:
      4
     /   
    2    
   /    
  3   1    

v = 1

d = 3

Output: 
      4
     /   
    2
   /     
  1   1
 /       
3       1

Note:

1. The given d is in range [1, maximum depth of the given tree + 1].
2. The given binary tree has at least one tree node.

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题意

在二叉树中的第d层前插入一层值全为v的结点，

思路

层序遍历找到第d-1层的所有结点，再插入新结点和修改老结点的左右子树即可。

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代码实现

Java

class Solution {
    public TreeNode addOneRow(TreeNode root, int v, int d) {
        if (root == null) return null;
        if (d == 1) return new TreeNode(v, root, null);

        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        int level = 1;

        while (level < d - 1) {
            int size = q.size();
            while (size > 0) {
                TreeNode cur = q.poll();
                if (cur.left != null) q.offer(cur.left);
                if (cur.right != null) q.offer(cur.right);
                size--;
            }
            level++;
        }

        while (!q.isEmpty()) {
            TreeNode cur = q.poll();
            cur.left = new TreeNode(v, cur.left, null);
            cur.right = new TreeNode(v, null, cur.right);
        }

        return root;
    }
}